NCERT Solutions for Class 11 Maths Chapter 15 - Statistics

Exercise 15.1: This exercise introduces the concept of statistics and its importance in real-life applications. Students will learn about the different types of data and how to organize data using frequency distribution tables.

Exercise 15.2: In this exercise, students will learn about measures of central tendency, including mean, median, and mode, and their properties. They will also practice finding the measures of central tendency for grouped data and ungrouped data.

Exercise 15.3: This exercise focuses on measures of dispersion, including range, quartile deviation, mean deviation, and standard deviation. Students will learn about the properties of these measures and practice finding them for grouped data and ungrouped data.

Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of statistics to solve various problems and answer questions. They will also practice finding measures of central tendency and dispersion for grouped data and ungrouped data, as well as organizing data using frequency distribution tables.

Access NCERT Solutions for Class 11 Maths  Chapter 15 – Statistics

Exercise 15.1

1. Find the mean deviation about the mean for the data \[{\text{4,}}\,{\text{7,}}\,{\text{8,}}\,{\text{9,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13,}}\,{\text{17}}\].

Ans: Consider the given data, which is, \[4,\,7,\,8,\,9,\,10,\,12,\,13,\,17\].

Therefore, the mean of the data is,

$\overline x  = \dfrac{{4 + 7 + 8 + 9 + 10 + 12 + 13 + 17}}{8} = \dfrac{{80}}{8} = 10$ 

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x $ can be calculated as, $ - 6,\, - 3, - 2,\, - 1,\,0,\,2,\,3,\,7$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $6,\,3,2,\,1,\,0,\,2,\,3,\,7$.

Now, the mean deviation about the mean is,

\[\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {\left| {{x_i} - \overline x } \right|} }}{8} = \dfrac{{6 + 3 + 2 + 1 + 0 + 2 + 3 + 7}}{8} = \dfrac{{24}}{8} = 3\] 

2. Find the mean deviation about the mean for the data \[{\text{38,}}\,{\text{70,}}\,{\text{48,40,}}\,{\text{42,}}\,{\text{55,}}\,{\text{63,}}\,{\text{46,}}\,{\text{54,}}\,{\text{44}}\].

Ans: Consider the given data, which is, \[38,\,70,\,48,40,\,42,\,55,\,63,\,46,\,54,\,44\].

Therefore, the mean of the data is,

$\overline x  = \dfrac{{38\, + 70 + \,48 + 40 + \,42 + \,55 + \,63 + \,46 + \,54 + \,44}}{{10}} = \dfrac{{500}}{{10}} = 50$ 

Observe that the deviations of the respective observations from the mean $\left( {\overline x } \right)$, that is, ${x_i} - \overline x $ can be calculated as, $ - 12,\,20, - 2,\, - 10,\, - 8,\,5,\,13,\, - 4,\,4,\, - 6$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - \overline x } \right|$ are $12,\,20,2,\,10,\,8,\,5,\,13,\,4,\,4,\,6$.

Now, the mean deviation about the mean is,

\[\therefore M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - \overline x } \right|} }}{{10}} = \dfrac{{12 + \,20 + 2 + \,10 + \,8 + \,5 + \,13 + \,4 + \,4 + \,6}}{{10}} = \dfrac{{84}}{{10}} = 8.4\] 

3. Find the mean deviation about the median for the data \[{\text{13,1}}\,{\text{7,}}\,{\text{16,14,}}\,{\text{11,}}\,{\text{13,}}\,{\text{10,}}\,{\text{16,}}\,{\text{11,}}\,{\text{18,}}\,{\text{12,}}\,{\text{17}}\].

Ans: Consider the given data, which is, \[13,1\,7,\,16,14,\,11,\,13,\,10,\,16,\,11,\,18,\,12,\,17\].

Observe that the number of observations in this case is $12$, that is, even and on arranging the data in ascending order it can be obtained as, \[10,1\,1,\,11,12,\,13,\,13,\,14,\,16,\,16,\,17,\,17,\,18\]  Therefore, the median of the data is the average of the ${6^{th}}$ and the ${7^{th}}$ observations,

$\therefore M = \dfrac{{13 + 14}}{2} = \dfrac{{27}}{2} = 13.5$ 

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $ - 3.5,\, - 2.5, - 2.5,\, - 1.5,\, - 0.5,\, - 0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $3.5,\,2.5,2.5,\,1.5,\,0.5,\,0.5,\,0.5,\,2.5,\,2.5,\,3.5,3.5,4.5$.

Now, the mean deviation about the median is,

\[\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{12} {\left| {{x_i} - M} \right|} }}{{12}} = \dfrac{{3.5 + \,2.5 + 2.5 + \,1.5 + \,0.5 + \,0.5 + \,0.5 + \,2.5 + \,2.5 + \,3.5 + 3.5 + 4.5}}{{12}}\]\[ \Rightarrow M.D.\left( M \right) = \dfrac{{28}}{{12}} = 2.33\] 

4. Find the mean deviation about the median for the data \[{\text{36,72,}}\,{\text{46,42,}}\,{\text{60,}}\,{\text{45,}}\,{\text{53,}}\,{\text{46,}}\,{\text{51,}}\,{\text{49}}\].

Ans: Consider the given data, which is, \[36,72,\,46,42,\,60,\,45,\,53,\,46,\,51,\,49\].

Observe that the number of observations in this case is $10$, that is, even and on arranging the data in ascending order it can be obtained as, \[36,42,\,45,46,\,46,\,49,\,51,\,53,\,60,\,72\]  Therefore, the median of the data is the average of the ${5^{th}}$ and the ${6^{th}}$ observations,

$\therefore M = \dfrac{{46 + 49}}{2} = \dfrac{{95}}{2} = 47.5$ 

Observe that the deviations of the respective observations from the median $\left( M \right)$, that is, ${x_i} - M$ can be calculated as, $ - 11.5,\, - 5.5, - 2.5,\, - 1.5,\, - 1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$ and therefore, the absolute value of the deviations calculated by $\left| {{x_i} - M} \right|$ are $11.5,\,5.5,2.5,\,1.5,\,1.5,\,1.5,\,3.5,\,5.5,\,12.5,\,24.5$.

Now, the mean deviation about the median is,

\[\therefore M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^{10} {\left| {{x_i} - M} \right|} }}{{10}} = \dfrac{{11.5 + \,5.5 + 2.5 + \,1.5 + \,1.5 + \,1.5 + \,3.5 + \,5.5 + \,12.5 + \,24.5}}{{10}}\]\[ \Rightarrow M.D.\left( M \right) = \dfrac{{70}}{{10}} = 7\] 

5. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}}  = \dfrac{1}{{25}} \times 350 = 14\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{158}}{{25}} = 6.32\]

6. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^5 {{f_i}{x_i}}  = \dfrac{1}{{80}} \times 4000 = 50\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1280}}{{80}} = 16\]

7. Find the mean deviation about the median for the data.

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

Observe that the number of observations in this case is $26$, that is, even and therefore, the median is the mean of the ${13^{th}}$ and the ${14^{th}}$ observations. Observe that both of these observations lie in the cumulative frequency $14$, for which the corresponding observation is obtained as $7$,

$\therefore M = \dfrac{{7 + 7}}{2} = \dfrac{{14}}{2} = 7$ 

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,  

Therefore, the mean deviation about the mean is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{16 + 4 + 6 + 10 + 48}}{{26}} = \dfrac{{84}}{{26}} = 3.23\]

8. Find the mean deviation about the median for the data.

Ans: It can be clearly observed that the given observations are already in ascending order and hence on adding a column corresponding to the cumulative frequencies of the given data, the following table can be obtained as shown below,

Observe that the number of observations, in this case, is $29$, which is, odd and therefore, the median is the ${15^{th}}$ observation. Observe that this observation lie in the cumulative frequency $21$, for which the corresponding observation is obtained as $30$,

$\therefore M = 30$ 

Now, the absolute values of the deviations from the median can be calculated using $\left| {{x_i} - M} \right|$ and therefore observe the table as shown below,  

Therefore, the mean deviation about the median is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^5 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{45 + 45 + 18 + 0 + 40}}{{29}} = \dfrac{{148}}{{29}} = 5.1\]

9. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^8 {{f_i}{x_i}}  = \dfrac{1}{{50}} \times 17900 = 358\] 

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{7896}}{{50}} = 157.92\]

10. Find the mean deviation about the mean for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^6 {{f_i}{x_i}}  = \dfrac{1}{{100}} \times 12530 = 125.3\]

Therefore, the mean deviation about the mean is,

\[M.D.\left( {\overline x } \right) = \dfrac{{\sum\limits_{i = 1}^6 {{f_i}\left| {{x_i} - \overline x } \right|} }}{N} = \dfrac{{1128.8}}{{100}} = 11.28\]

11. Calculate the mean deviation about the median age for the age distribution of ${\text{100}}$ persons.

Ans: It can be clearly observed that the given data is not continuous and therefore it needs to be converted into a continuous frequency distribution, which can be done by subtracting $0.5$ from the lower limit and adding $0.5$ to the upper limit of each class interval. Now, observe the table as shown below, which is,

Observe that the class interval containing the ${\left( {\dfrac{N}{2}} \right)^{th}}$ item or the ${50^{th}}$ item is $35.5 - 40.5$. Thus, $35.5 - 40.5$ is the median class.

Therefore, median of the data can be calculated as shown below,

\[\therefore M = l + \dfrac{{\dfrac{N}{2} - C}}{f} \times h\]   [ where $l = 35.5$, $C = 37$, $f = 26$, $h = 5$ and $N = 100$] 

\[ \Rightarrow M = 35.5 + \dfrac{{50 - 37}}{{26}} \times 5 = 35.5 + \dfrac{{13}}{{26}} \times 5 = 35.5 + 2.5 = 38\]

Therefore, the mean deviation about the median is,

\[M.D.\left( M \right) = \dfrac{{\sum\limits_{i = 1}^8 {{f_i}\left| {{x_i} - M} \right|} }}{N} = \dfrac{{735}}{{100}} = 7.35\]

Exercise 15.2

1. Find the mean and variance for the data \[{\text{6,}}\,{\text{7,}}\,{\text{10,}}\,{\text{12,}}\,{\text{13, 4,}}\,{\text{8,}}\,{\text{12}}\].

Ans: Consider the given data which is, \[6,\,7,\,10,\,12,\,13,{\text{ }}4,\,8,\,12\].

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^8 {{x_i}}  = \dfrac{{6 + 7 + 10 + 12 + 13 + 4 + 8 + 12}}{8} = \dfrac{{72}}{8} = 9\] 

 Now, observe the table as shown below, which is,

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n} = \dfrac{{74}}{8} = 9.25\]

2. Find the mean and variance for the first ${\text{n}}$ natural numbers.

Ans: The mean of the first $n$ natural numbers can be calculated as shown below,

\[\therefore \overline x  = \dfrac{{\dfrac{{n\left( {n + 1} \right)}}{2}}}{n} = \dfrac{{n + 1}}{2}\] 

Now, the variance can be calculated as,

\[\therefore Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {\left[ {{x_i} - {{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \right]} \]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2}  - \dfrac{1}{n}\sum\limits_{i = 1}^n {2\left( {\dfrac{{n + 1}}{n}} \right){x_i}}  + \dfrac{1}{n}\sum\limits_{i = 1}^n {{{\left( {\dfrac{{n + 1}}{2}} \right)}^2}} \]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{\left( {n + 1} \right)(2n + 1)}}{6} - \dfrac{{{{\left( {n + 1} \right)}^2}}}{2} + \dfrac{{{{\left( {n + 1} \right)}^2}}}{4}\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \left( {n + 1} \right)\left[ {\dfrac{{4n + 2 - 3n - 3}}{{12}}} \right]\]

\[ \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{{n^2} - 1}}{{12}}\]

3. Find the mean and variance for the first ${\text{10}}$ multiples of ${\text{3}}$.

Ans: Observe that the first $10$ multiples of $3$ are, \[3,\,6,\,9,\,12,\,15,\,18,\,21,\,24,\,27,\,30\].

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^{10} {{x_i}}  = \dfrac{{3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 + 30}}{{10}} = \dfrac{{165}}{{10}} = 16.5\] 

 Now, observe the table as shown below, which is,

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n} = \dfrac{{742.5}}{{10}} = 74.25\]

4. Find the mean and variance for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}}  = \dfrac{1}{{40}} \times 760 = 19\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{1736}}{{40}} = 43.4\]

5. Find the mean and variance for the data.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^7 {{f_i}{x_i}}  = \dfrac{1}{{22}} \times 2200 = 100\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{640}}{{22}} = 29.09\]

6. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 64 + \dfrac{0}{{100}} \times 1 = 64\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{1}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 286 - 0} \right] = 2.86\]

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {2.86}  = 1.69\]

7. Find the mean and variance for the following frequency distribution.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h = 105 + \dfrac{2}{{30}} \times 30 = 107\] 

Therefore, the variance is,

\[Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {30} \right)}^2}}}{{{{\left( {30} \right)}^2}}}\left[ {30 \times 76 - {{\left( 2 \right)}^2}} \right] = 2280 - 4 = 2276\]

8. Find the mean and variance for the following frequency distribution.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 25 + \dfrac{{10}}{{50}} \times 10 = 27\] 

Therefore, the variance is,

$ Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - {{\left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)}^2}} } \right] = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {50} \right)}^2}}}\left[ {50 \times 68 - {{\left( {10} \right)}^2}} \right] \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{1}{{25}}\left[ {3400 - 100} \right] = 132  \\ $

9. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^9 {{f_i}{y_i}} }}{N} \times h = 92.5 + \dfrac{6}{{60}} \times 5 = 92.5 + 0.5 = 93\] 

Therefore, the variance is,

$Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^9 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^9 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 5 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {60 \times 254 - {{\left( 6 \right)}^2}} \right]  \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{25}}{{3600}}\left( {15204} \right) = 105.58  \\ $

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {105.58}  = 10.27\]

10. The diameters of circles (in mm) drawn in a design are given below. Find the mean, variance and standard deviation using the shortcut method.

Ans: Consider the given data and observe the table as shown below, which is,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^5 {{f_i}{y_i}} }}{N} \times h = 42.5 + \dfrac{{25}}{{100}} \times 4 = 43.5\] 

Therefore, the variance is,

$  Var\left( {{\sigma ^2}} \right) = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^5 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^5 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( 4 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\left[ {100 \times 199 - {{\left( {25} \right)}^2}} \right]  \\$

$   \Rightarrow Var\left( {{\sigma ^2}} \right) = \dfrac{{16}}{{10000}}\left( {19900 - 625} \right) = \dfrac{{16}}{{10000}}\left( {19275} \right) = 30.84 \\\ $

Now the standard deviation can be calculated as shown below,

\[\therefore \sigma  = \sqrt {30.84}  = 5.55\]

Exercise-15.3

1. From the data given below state which group is more variable, A or B? 

Ans: Consider the given data and observe the table for Group A as shown below,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h = 45 + \dfrac{{ - 6}}{{150}} \times 10 = 45 - 0.4 = 44.6\] 

Now the standard deviation can be calculated as shown below,

$\therefore {\sigma _A}^2 = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {150} \right)}^2}}}\left[ {150 \times 342 - {{\left( { - 6} \right)}^2}} \right] \\$

$   \Rightarrow {\sigma _A} = \sqrt {\dfrac{1}{{225}}\left( {51264} \right)}  = \sqrt {227.84}  = 15.09 \\ $

Again, consider the given data and observe the table for Group A as shown below,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = A + \dfrac{{\sum\limits_{i = 1}^7 {{f_i}{y_i}} }}{N} \times h = 45 + \dfrac{{ - 6}}{{150}} \times 10 = 45 - 0.4 = 44.6\] 

Now the standard deviation can be calculated as shown below,

$\therefore {\sigma _A}^2 = \dfrac{{{h^2}}}{{{N^2}}}\left[ {N{{\sum\limits_{i = 1}^7 {{f_i}{y_i}^2 - \left( {\sum\limits_{i = 1}^7 {{f_i}{y_i}} } \right)} }^2}} \right] = \dfrac{{{{\left( {10} \right)}^2}}}{{{{\left( {150} \right)}^2}}}\left[ {150 \times 366 - {{\left( { - 6} \right)}^2}} \right]  \\$

 $  \Rightarrow {\sigma _A} = \sqrt {\dfrac{1}{{225}}\left( {54864} \right)}  = \sqrt {243.84}  = 15.61 \\ $

It is clearly known that as the mean of both the groups is the same, then the group with greater s

standard deviation will be more variable. Henceforth, group B has more variability in the marks.

2. From the price of shares of ${\text{X}}$ and ${\text{Y}}$ below, find out which is more stable in value.

Ans: Observe that the prices of the shares of $X$ are, \[35,\,54,\,52,\,53,\,56,\,58,\,52,\,50,\,51,\,49\].

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{x_i}}  = \dfrac{{510}}{{10}} = 51\] 

 Now, observe the table as shown below for shares $X$ , which is,

Therefore, the standard deviation is,

\[{\sigma _X}^2 = \dfrac{{\sum\limits_{i = 1}^{10} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{350}}{{10}} = 35\]

\[ \Rightarrow {\sigma _X} = \sqrt {35}  = 5.91\]

Now the coefficient of variation of price of shares $X$ can be calculated as shown below,

\[\therefore C.V{._X} = \dfrac{{{\sigma _X}}}{{\overline x }} \times 100 = \dfrac{{5.91}}{{51}} \times 100 = 11.58\]

Again, observe that the prices of the shares of $Y$ are, \[108,\,107,\,105,\,105,\,106,\,107,\,104,\,103,\,104,\,101\].

The mean of the data can be calculated as shown below,

\[\therefore \overline y  = \dfrac{1}{N}\sum\limits_{i = 1}^{10} {{y_i}}  = \dfrac{{1050}}{{10}} = 105\] 

 Now, observe the table as shown below for shares $Y$ , which is,

Therefore, the standard deviation is,

\[{\sigma _Y}^2 = \dfrac{{\sum\limits_{i = 1}^{10} {{{\left( {{y_i} - \overline y } \right)}^2}} }}{N} = \dfrac{{40}}{{10}} = 4\]

\[ \Rightarrow {\sigma _Y} = \sqrt 4  = 2\]

Now the coefficient of variation of price of shares $Y$ can be calculated as shown below,

\[\therefore C.V{._Y} = \dfrac{{{\sigma _Y}}}{{\overline y }} \times 100 = \dfrac{2}{{105}} \times 100 = 1.9\]

It can be observed that $C.V{._X}$ is greater than $C.V{._Y}$ and henceforth the price of shares of $Y$ are more stable than the price of shares $X$.

3. An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:  

(i) Which firm A or B pays larger amount as monthly wages? 

Ans: Observe that the mean of monthly wages for firm A is $Rs.\,5253$ and the total number of wage earners in firm A is $586$. Therefore, the total amount of monthly wages paid in firm A is,

$ = Rs.\,5253 \times 586$ 

Again, observe that the mean of monthly wages for firm B is $Rs.\,5253$ and the total number of wage earners in firm B is $648$. Therefore, the total amount of monthly wages paid in firm A is,

$ = Rs.\,5253 \times 648$ 

Henceforth, firm B pays a larger amount of monthly wages when compared to firm A as the total number of wage earners in firm B is higher than in case of firm A.

(ii) Which firm A or B pays larger amount as monthly wages? 

Ans: It is given that the variance of the distribution of wages in firm A is $100$.

Now the standard deviation of the distribution of wages can be calculated as shown below,

\[\therefore {\sigma _A} = \sqrt {100}  = 10\] 

Again, it is given that the variance of the distribution of wages in firm B is $121$.

Now the standard deviation of the distribution of wages can be calculated as shown below,

\[\therefore {\sigma _B} = \sqrt {121}  = 11\] 

Observe that the mean of monthly wages of both the firms is same, which is, $Rs.\,5253$. Thus, the firm with greater standard deviation will have more variability and henceforth, firm B has got more variability in the individual wages.

4. The following is the record of goals scored by team A in a football session: 

For the team B, mean number of goals scored per match was ${\text{2}}$ with a standard deviation ${\text{1}}{\text{.25}}$ goals. Find which team may be considered more consistent?  

Ans: Calculate the mean and the standard deviation of goals scored by team A as shown below,

The mean of the data can be calculated as shown below,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{15} {{f_i}{x_i}} }}{N} = \dfrac{{50}}{{25}} = 2\] 

It is clearly known that the mean of both the teams is same.

Now the standard deviation can be calculated as shown below,

\[\therefore {\sigma _A} = \dfrac{1}{N}\sqrt {N{{\sum\limits_{i = 1}^{15} {{f_i}{x_i}^2 - \left( {\sum\limits_{i = 1}^{15} {{f_i}{x_i}} } \right)} }^2}} \]

\[ \Rightarrow {\sigma _A} = \dfrac{1}{{25}}\sqrt {25 \times 130 - {{\left( {50} \right)}^2}}  = \dfrac{1}{{25}}\sqrt {750}  = 1.09\]

The standard deviation of team B is clearly given as $1.25$ goals. Observe that the mean number of goals scored by both the teams is same, which is, $2$. Thus, the firm with lower standard deviation will be more consistent and henceforth, team A has got lower standard deviation and is more consistent when compared to team B.

5. The sum and sum of squares corresponding to the length ${\text{x}}$ (in cm) and weight ${\text{y}}$ (in gm) of ${\text{50}}$ plant products are given below:

\[\sum\limits_{{\text{i = 1}}}^{{\text{50}}} {{{\text{x}}_{\text{i}}}} {\text{ = 212,}}\,\sum\limits_{{\text{i = 1}}}^{{\text{50}}} {{{\text{x}}_{\text{i}}}^{\text{2}}} {\text{ = 902}}{\text{.8,}}\,\sum\limits_{{\text{i = 1}}}^{{\text{50}}} {{{\text{y}}_{\text{i}}}} {\text{ = 261,}}\,\sum\limits_{{\text{i = 1}}}^{{\text{50}}} {{{\text{y}}_{\text{i}}}^{\text{2}}} {\text{ = 1457}}{\text{.6}}\] 

Which is more varying the length or the weight?

Ans: Observe that the given information is, \[\sum\limits_{i = 1}^{50} {{x_i}}  = 212,\,\sum\limits_{i = 1}^{50} {{x_i}^2}  = 902.8,\,\sum\limits_{i = 1}^{50} {{y_i}}  = 261,\,\sum\limits_{i = 1}^{50} {{y_i}^2}  = 1457.6\] where $x$ represents the length (in cm) and $y$ represents the weight (in gm) of $50$ plant products 

Therefore, the mean of the lengths can be calculated as shown below,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{50} {{x_i}} }}{N} = \dfrac{{50}}{{25}} = 2\] 

Now the standard deviation can be calculated as shown below,

\[{\sigma _1}^2 = \dfrac{{\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = \dfrac{{\sum\limits_{i = 1}^{50} {{{\left( {{x_i} - 4.24} \right)}^2}} }}{{50}}\]

\[ \Rightarrow {\sigma _1}^2 = \dfrac{1}{{50}}\left[ {\sum\limits_{i = 1}^{50} {{x_i}^2}  - 8.48\sum\limits_{i = 1}^{50} {{x_i}}  + \left( {17.97 \times 50} \right)} \right]\]

\[ \Rightarrow {\sigma _1}^2 = \dfrac{1}{{50}}\left[ {902.8 - \left( {8.48 \times 212} \right) + 898.5} \right]\]

\[ \Rightarrow {\sigma _1} = \sqrt {\dfrac{1}{{50}}\left[ {902.8 - \left( {8.48 \times 212} \right) + 898.5} \right]}  = \sqrt {0.07}  = 0.26\]

Now the coefficient of variation of lengths can be calculated as shown below,

\[\therefore C.V{._1} = \dfrac{{{\sigma _1}}}{{\overline x }} \times 100 = \dfrac{{0.26}}{{4.24}} \times 100 = 6.13\]

Again, the mean of the weights can be calculated as shown below,

\[\therefore \overline y  = \dfrac{{\sum\limits_{i = 1}^{50} {{y_i}} }}{N} = \dfrac{{261}}{{50}} = 5.22\] 

Now the standard deviation can be calculated as shown below,

\[{\sigma _2}^2 = \dfrac{{\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - \overline y } \right)}^2}} }}{N} = \dfrac{{\sum\limits_{i = 1}^{50} {{{\left( {{y_i} - 5.22} \right)}^2}} }}{{50}}\]

\[ \Rightarrow {\sigma _2}^2 = \dfrac{1}{{50}}\left[ {\sum\limits_{i = 1}^{50} {{y_i}^2}  - 10.44\sum\limits_{i = 1}^{50} {{y_i}}  + \left( {27.24 \times 50} \right)} \right]\]

\[ \Rightarrow {\sigma _2}^2 = \dfrac{1}{{50}}\left[ {1457.6 - \left( {10.44 \times 261} \right) + 1362} \right]\]

\[ \Rightarrow {\sigma _2} = \sqrt {\dfrac{1}{{50}}\left[ {1457.6 - \left( {10.44 \times 261} \right) + 1362} \right]}  = \sqrt {1.89}  = 1.37\]

Now the coefficient of variation of weights can be calculated as shown below,

\[\therefore C.V{._2} = \dfrac{{{\sigma _2}}}{{\overline y }} \times 100 = \dfrac{{1.37}}{{5.22}} \times 100 = 26.24\]

It can be observed that $C.V.$ of weights is greater than $C.V.$ of lengths and henceforth the weights vary more than the lengths.

Miscellaneous Exercise

1. The mean and variance of eight observations are ${\text{9}}$ and ${\text{9}}{\text{.25}}$ respectively. If six of the observations are \[{\text{6,}}\,{\text{7,10,}}\,{\text{12,}}\,{\text{12}}\] and \[{\text{13}}\], find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[6,\,7,10,\,12,\,12,\,13,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{6 + 7 + \,10 + 12 + \,12 + 13 + x + y}}{8} = 9\] 

\[ \Rightarrow 60 + x + y = 72\]

\[ \Rightarrow x + y = 12\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^8 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 9.25\]

\[\therefore \dfrac{1}{8}\left[ {{{\left( { - 3} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {x^2} + {y^2} - (2)(9)\left( {x + y} \right) + \left( {2{{\left( 9 \right)}^2}} \right)} \right] = 9.25\]

\[ \Rightarrow \dfrac{1}{8}\left[ {9 + 4 + 9 + 16 + {x^2} + {y^2} - (18)\left( {12} \right) + 162} \right] = 9.25\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{8}\left[ {{x^2} + {y^2} - 6} \right] = 9.25\]

\[ \Rightarrow {x^2} + {y^2} = 80\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 144$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 64\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 16\] 

\[ \Rightarrow x - y =  \pm 4\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 4$, when \[x - y = 4\] and $x = 4$ and $y = 8$, when \[x - y =  - 4\]. Henceforth, the remaining two observations are $4$ and $8$.

2. The mean and variance of ${\text{7}}$ observations are ${\text{8}}$ and ${\text{16}}$ respectively. If five of the observations are \[{\text{2,}}\,{\text{4,10,}}\,{\text{12}}\] and \[{\text{14}}\]. Find the remaining two observations.

Ans: Consider the remaining two observations to be $x$ and $y$. Thus the eight observations are \[2,\,4,10,\,12,\,14,\,x,\,y\].

Therefore, from the mean of the data it can be obtained that,

\[\overline x  = \dfrac{{2 + 4 + \,10 + 12 + \,14 + x + y}}{7} = 8\] 

\[ \Rightarrow 42 + x + y = 56\]

\[ \Rightarrow x + y = 14\] ......(i)

Again, from the variance of the data it can be obtained that,

\[{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^7 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{N} = 16\]

\[\therefore \dfrac{1}{7}\left[ {{{\left( { - 6} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 6 \right)}^2} + {x^2} + {y^2} - (2)(8)\left( {x + y} \right) + \left( {2{{\left( 8 \right)}^2}} \right)} \right] = 16\]

\[ \Rightarrow \dfrac{1}{7}\left[ {36 + 16 + 4 + 16 + {x^2} + {y^2} - (16)\left( {14} \right) + 128} \right] = 16\]   [By, using (i)]

\[ \Rightarrow \dfrac{1}{7}\left[ {{x^2} + {y^2} + 12} \right] = 16\]

\[ \Rightarrow {x^2} + {y^2} = 100\] ......(ii)

Observe that form equation (i), it can be obtained that,

${x^2} + {y^2} + 2xy = 196$ ......(iii)

Also, from equations (ii) and (iii), it can be obtained that,

\[2xy = 96\] ......(iv) 

Now, on subtracting equation (iv) from equation (ii), it can be obtained that,

\[{x^2} + {y^2} - 2xy = 4\] 

\[ \Rightarrow x - y =  \pm 2\] ......(v)

It can be calculated from equations (i) and (v) that, $x = 8$ and $y = 6$, when \[x - y = 2\]and $x = 6$ and $y = 8$, when \[x - y =  - 2\]. Henceforth, the remaining two observations are $6$ and $8$.

3. The mean and standard deviation of six observations are ${\text{8}}$ and ${\text{4}}$ respectively. If each observation is multiplied by ${\text{3}}$, find the new mean and new standard deviation of the resulting observations.

Ans: Assume the observations to be \[{x_1},\,{x_2},\,{x_3},\,{x_4},\,{x_5},\,{x_6}\].

Therefore, the mean of the data is,

\[\overline x  = \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] 

\[ \Rightarrow \dfrac{{{x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6}}}{6} = 8\] ......(i)

Observe that when each of the observation is multiplied with $3$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_1} = 3{x_1}$ 

$ \Rightarrow {x_1} = \dfrac{1}{3}{y_1},\,\forall i = 1,\,2,\,3,\,....,\,6\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

\[\therefore \overline y  = \dfrac{{{y_1} + {y_2} + {y_3} + {y_4} + {y_5} + {y_6}}}{6}\] 

\[ \Rightarrow \overline y  = \dfrac{{3({x_1} + {x_2} + {x_3} + {x_4} + {x_5} + {x_6})}}{6}\]

\[ \Rightarrow \overline y  = 3 \times 8\]      [By using (i)]

\[ \Rightarrow \overline y  = 24\]

Therefore, the standard deviation of the data is,

\[\sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}}  = 4\]

\[ \Rightarrow {\left( 4 \right)^2} = \dfrac{{\sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}} }}{6}\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{x_i} - \overline x } \right)}^2}}  = 96\] ......(ii)

Again, it can be observed from equations (i) and (ii) that \[\overline y  = 3\overline x \] and \[\overline x  = \dfrac{{\overline y }}{3}\] and hence on substituting the values of ${x_i}$ and $\overline x $ in equation (ii) it can be clearly obtained as shown below,

 \[\therefore \sum\limits_{i = 1}^6 {{{\left( {\dfrac{{{y_i}}}{3} - \dfrac{{\overline y }}{3}} \right)}^2}}  = 96\]

\[ \Rightarrow \sum\limits_{i = 1}^6 {{{\left( {{y_i} - \overline y } \right)}^2}}  = 864\]

Henceforth, the standard deviation of the new data can be calculated as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{{864}}{6}}  = \sqrt {144}  = 12\]

4. Given that $\overline {\text{x}} $ is the mean and $\sigma^{2}$ is the variance of ${\text{n}}$ observations ${{\text{x}}_{\text{1}}}{\text{,}}\,{{\text{x}}_{\text{2}}}{\text{,}}\,......{\text{,}}\,{{\text{x}}_{\text{n}}}$.Prove that the mean and variance of observations ${\text{a}}{{\text{x}}_{\text{1}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{2}}}{\text{,}}\,{\text{a}}{{\text{x}}_{\text{3}}}{\text{, }}.....{\text{,}}\,{\text{a}}{{\text{x}}_{\text{n}}}$ are ${\text{a}}\overline {\text{x}} $ and ${{\text{a}}^{\text{2}}}{{\sigma}}^{\text{2}}$, respectively $\left( {{\text{a}} \ne {\text{0}}} \right)$.

Ans: Observe that the given observations are ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{y_i}{{\left( {{x_i} - \overline x } \right)}^2}} }}{n}\]......(i)

Observe that when each of the observation is multiplied with $a$ and if we consider the resulting observations as ${y_i}$, then observe as shown below,

$\therefore {y_i} = a{x_i}$ 

$ \Rightarrow {x_i} = \dfrac{1}{a}{y_i},\,\forall i = 1,\,2,\,3,\,....,\,n\& i \in {\mathbb{Z}^ + }$  

Now, the mean of the new data is,

$\therefore \overline y  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{y_i}}  = \dfrac{1}{n}\sum\limits_{i = 1}^n {a{x_i}}  = \dfrac{a}{n}\sum\limits_{i = 1}^n {{x_i}}  = a\overline x \,\,\,\,\,\,\,\left[ {\because \overline x  = \dfrac{1}{n}\sum\limits_{i = 1}^n {{x_i}} } \right]$ 

Again on substituting the values of ${x_i}$ and $\overline x $ in equation (i) it can be clearly obtained as shown below,

 \[\therefore {\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {\dfrac{{{y_i}}}{a} - \dfrac{{\overline y }}{a}} \right)}^2}} }}{n}\]

\[ \Rightarrow {a^2}{\sigma ^2} = \dfrac{{\sum\limits_{i = 1}^n {{{\left( {{y_i} - \overline y } \right)}^2}} }}{n}\]

Henceforth, it can be clearly proved that the mean and variance of the new data is $a\overline x $ and ${a^2}{\sigma ^2}$, respectively.

5. The mean and standard deviation of ${\text{20}}$ observations are found to be ${\text{10}}$ and ${\text{2}}$, respectively. On rechecking it was found that an observation ${\text{8}}$ was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

(i) If wrong item is omitted 

Ans: Observe that the incorrect number of observations, incorrect mean and the incorrect standard deviation are $20$, $10$ and $2$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n} = 10\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{{20}} = 10\] 

\[ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}}  = 200\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{19} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{192}}{{19}} = 10.1\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{19} {{x_i}}  = 192} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2}\]

\[ \Rightarrow 2080 - 64\]

\[ \Rightarrow 2016\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{19} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2016}}{{19}} - {{\left( {10.1} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {1061.1 - 102.1}  = \sqrt {4.09}  = 2.02\]

(ii) If it is replaced by ${\text{12}}$.  

Ans: Observe that the incorrect sum of observations is $200$.

\[\therefore \sum\limits_{i = 1}^{20} {{x_i}}  = 200 - 8 + 12 = 204\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{20} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{204}}{{20}} = 10.2\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{20} {{x_i}}  = 204} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 2\]

\[ \Rightarrow 4 = \dfrac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 2080 = \sum\limits_{i = 1}^{20} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{20} {{x_i}^2}  - {\left( 8 \right)^2} + {\left( {12} \right)^2}\]

\[ \Rightarrow 2080 - 64 + 144\]

\[ \Rightarrow 2160\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{20} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{2160}}{{20}} - {{\left( {10.2} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {108 - 104.04}  = \sqrt {3.96}  = 1.98\]

6.The mean and standard deviation of marks obtained by ${\text{50}}$ students of a class in three subjects, Mathematics, Physics and Chemistry are given below: 

Which of these three subjects shows the highest variability in marks and which shows the lowest? 

Ans: Observe that the standard deviation of the three subjects, that is, Mathematics and Physics and Chemistry are $12,\,15$ and $20$ respectively. 

Now the coefficient of variation of the three subjects can be calculated as shown below,

\[\therefore C.V{._M} = \dfrac{{{\sigma _M}}}{{{{\overline x }_M}}} \times 100 = \dfrac{{12}}{{42}} \times 100 = 28.57\]

\[\therefore C.V{._P} = \dfrac{{{\sigma _P}}}{{{{\overline x }_P}}} \times 100 = \dfrac{{15}}{{32}} \times 100 = 46.87\]

\[\therefore C.V{._C} = \dfrac{{{\sigma _C}}}{{{{\overline x }_C}}} \times 100 = \dfrac{{20}}{{40.9}} \times 100 = 48.89\]

It is clearly known that the subject with greater $C.V.$ is more variable than the others and henceforth it can be observed that $C.V.$ of Chemistry is greater than $C.V.$ of Mathematics and Physics and $C.V.$ of Mathematics is lower than $C.V.$ of Chemistry and Physics. Therefore, the highest variability of marks occur in Chemistry and the lowest variability of marks occur in Mathematics.

7. The mean and standard deviation of a group of ${\text{100}}$ observations were found to be ${\text{20}}$ and ${\text{3}}$, respectively. Later on it was found that ${\text{3}}$ observations are incorrect which were recorded as ${\text{21,}}\,{\text{21}}$ and ${\text{18}}$. Find the mean and standard deviation if the incorrect  observations are omitted.

Ans: Observe that the number of observations, incorrect mean and the incorrect standard deviation are $100$, $20$ and $3$, respectively. ${x_1},\,{x_2},\,......,\,{x_n}$ and the mean and variance of data is $\overline x $ and ${\sigma ^2}$ respectively.

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{n} = 20\] 

\[ \Rightarrow \dfrac{{\sum\limits_{i = 1}^{100} {{x_i}} }}{{100}} = 20\] 

\[ \Rightarrow \sum\limits_{i = 1}^{100} {{x_i}}  = 2000\]

Now, the correct mean is,

\[\therefore \overline x  = \dfrac{{\sum\limits_{i = 1}^{97} {{x_i}} }}{n}\] 

\[ \Rightarrow \overline x  = \dfrac{{2000 - 60}}{{97}} = 20\,\,\,\,\,\left[ {\because \sum\limits_{i = 1}^{97} {{x_i}}  = 1940} \right]\]

Again observe as shown below,

\[\therefore \sigma  = \sqrt {\dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} }  = 3\]

\[ \Rightarrow 9 = \dfrac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}^2 - {{\left( {\overline x } \right)}^2}} \]

\[ \Rightarrow 40900 = \sum\limits_{i = 1}^{100} {{x_i}^2} \]

Therefore, the correct standard deviation of the data is,

\[\sum\limits_{i = 1}^{100} {{x_i}^2}  - (2){\left( {21} \right)^2} - {(18)^2}\]

\[ \Rightarrow 40900 - 1206\]

\[ \Rightarrow 39694\]

\[\therefore \sigma  = \sqrt {\dfrac{{\sum\limits_{i = 1}^{97} {{x_i}^2} }}{n} - {{\left( {\overline x } \right)}^2}} \] 

\[ \Rightarrow \sigma  = \sqrt {\dfrac{{39694}}{{97}} - {{\left( {20} \right)}^2}} \]

\[ \Rightarrow \sigma  = \sqrt {409.22 - 400}  = \sqrt {9.22}  = 3.04\].

Class 11 Maths NCERT Solutions Chapter 15

Class 11 Maths Ch 15 NCERT Solutions shows the sums on a step-by-step basis. The concepts touched up in the chapter and the example sums are indicated below:

Mean Deviation About Mean

1. Ungrouped

Example: Finding mean deviation about the mean of 6, 7, 10, 12, 13, 12, 8, 4

2. Discrete Frequency

Example: Finding mean deviation about the mean of the below data:

3. Continuous Frequency Distribution

Example: Finding the mean deviation about the mean of the below data:

• Mean Deviation About Median

1. Ungrouped

Example: Finding the mean deviation about the median of 3, 5, 9, 3, 10, 12, 4, 18, 7, 19, 21

2. Discrete Frequency

Example: Finding mean deviation about the median of the below data:

3. Continuous Frequency Distribution

Example: Calculation of the mean deviation about median of the below data:

• Standard Deviation and Variance

1. Ungrouped Data

Example: Finding the variance of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24

2. Discrete Frequency

Example: Finding the variance and standard deviation of the below data:

3. Continuous Frequency

Example: Calculation of the mean, variance and standard deviation of the following distribution: 

• Coefficient of Variation

Example: Determine which group is more variable from the below data:

• Indirect Questions

1. Multiplication of Observation

Example: Given that the mean and standard deviation of six observations are 8 and 4 respectively, if each observation is multiplied by 3, what would be the new mean and new standard deviation of the resultant observations?

2. Finding Remaining Observations

Example: The variance and mean of 7 observations are 16 and 8 respectively. If 5 of the observations amount to 2, 4, 10, 12, 14, what are the remaining 2 observations?

3. Incorrect Observation

Example: The standard deviation and mean of 100 observations are 5.1 and 40 respectively. However, a mistake was detected in case of one observation where instead of 40, 50 was taken. What would be the correct standard deviation and mean?

How is Statistics Class 11 Maths NCERT PDF Advantageous to Students?

Preparation of Class 11 Maths Statistics requires a higher degree of dedication and diligence from students. In that respect, Class 11 Maths Statistics solutions facilitates the process for students. Here are few of the prominent reasons why the solutions are advantageous:

1. In-Depth Content With Lucid Language: Class 11 Statistics solutions discuss the topics in an elaborate manner, and at the same ensures that it is done in a simple language.

2. Follow Approved Curriculum: Class 11 Maths chapter 15 solutions strictly adheres to the CBSE curriculum. Hence, these solutions itself form the base for board examination preparation. 

3. Explanation of Fundamental Concepts: Before proceeding on to advanced topics, basic concepts are dealt with in an exhaustive manner so that doubts or confusion of students may get clarified.

So, wait no more and download NCERT solutions for Class 11 Maths Chapter 15 Statistics from Vedantu portal now!

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 13 - Limits and Derivatives

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability

NCERT Solution Class 11 Maths of Chapter 15 All Exercises